$27 g m$ of $A l$ react with an excess of oxygen to give $4.59 g m$ of $A l_{2} O_{3}$ . Calculate the percentage yield of the reaction.

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Solution

$2 A l + \frac{3}{2} O_{2} ⟶ A l_{2} O_{3}$
Moles of $A l = \frac{27}{27} = 1$ mole
Moles of $A l_{2} O_{3} = \frac{4.59}{102} = 0.045$
By stiochiometry, $1$ mole $A l_{2} O_{3}$ requires $2$ moles of $A l$
$∴$ $0.045$ moles $A l_{2} O_{3}$ will require 0.09 moles of $A l$
$∴$ Moles of $A l$ reacted= $0.09$
$∴$ % yield= $\frac{m o l e s o f A l r e a c t e d}{m o l e s o f A l f e d} \times 100$
= $\frac{0.09}{1} \times 100$
$∴$ % yield= $9$ %
$∴$ Percentage yield of reaction is $9$ %

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